Mo2men wrote:

GMATGuruNY

Can you please show how I can open the modulus in statement 2 instead of squaring both sides .

Thanks

Statement 2: |x-y| < |x|

Case 1: x>0 and x≥y

In this case,

|x-y| = x-y and |x| = x.

Substituting the blue equations into |x-y| < x, we get:

x-y < x

-y < 0

y > 0

Implication:

If x>0 and x≥y, then y>0.

Thus:

0<y≤x

Case 2: x>0 and x<y

In this case,

|x-y| = -x+y and |x| = x.

Substituting the blue equations into |x-y| < x, we get:

-x+y < x

y < 2x

Implication:

If x>0 and x<y, then y<2x.

Thus:

0<x<y<2x

Case 3: x<0 and x≥y

In this case,

|x-y| = x-y and |x| = -x.

Substituting the blue equations into |x-y| < x, we get:

x-y < -x

2x < y

Implication:

If x<0 and x≥y, then y>2x.

Thus:

2x<y≤x<0

Case 4: x<0 and x<y

In this case,

|x-y| = -x+y and |x| = -x.

Substituting the blue equations into |x-y| < x, we get:

-x+y < -x

y < 0

Implication:

If x<0 and x<y, then y<0.

Thus:

x<y<0

In all four cases, x and y have the same sign, so xy>0.

I do not recommend this approach.

Squaring the inequality seems easier and faster.

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